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Is ! even? (1) E=Q (2) Q=2

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Is ! even? (1) E=Q (2) Q=2 [#permalink] New post 18 Dec 2004, 17:09
Is [E!+Q!]! even?

(1) E=Q
(2) Q=2
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 [#permalink] New post 18 Dec 2004, 17:57
I originally thought C, but I changed my mind. I think it's A. I thought originally that we had to be sure they weren't 0 or 1, because either of those factorial make 1, and 1 is odd. But knowing that the variables are equal means that even if they are 0 or 1, they both make 1, and 1 + 1 is even. So that's enough.

In 2, knowing that one variable is 2 doesn't tell us about the other one, which could be 0 or 1. So in either case, we'd get 3, which is odd, but in every other case we'd have even plus even which is even (every number other than 0 or 1 factorial must be even since it has at a minimum 2 times something in it).

So I'm going with A.
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 [#permalink] New post 18 Dec 2004, 21:35
ian7777 wrote:
I originally thought C, but I changed my mind. I think it's A. I thought originally that we had to be sure they weren't 0 or 1, because either of those factorial make 1, and 1 is odd. But knowing that the variables are equal means that even if they are 0 or 1, they both make 1, and 1 + 1 is even. So that's enough.

In 2, knowing that one variable is 2 doesn't tell us about the other one, which could be 0 or 1. So in either case, we'd get 3, which is odd, but in every other case we'd have even plus even which is even (every number other than 0 or 1 factorial must be even since it has at a minimum 2 times something in it).

So I'm going with A.

nice explanation Ian
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 [#permalink] New post 18 Dec 2004, 21:49
the only odd factorial is 0!=1!=1, all the rest are even.
Still A?
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 [#permalink] New post 18 Dec 2004, 22:14
I would go with D.

Infact, from the stem, it looks like it will never be an odd number. You will never get 0! or 1! from E!+Q! :?:
Besides 1! factorial every other factorial is multiplied by atleast 2 an even number and any integer multiplied by an even number will always be even.
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 [#permalink] New post 19 Dec 2004, 01:09
yeah. in fact, I think it would be even no matter what. I didn't see the exclamation point after the absolute value signs originally. I saw it without that, so there was the chance of getting 2+1=3.
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Re: more fac [#permalink] New post 19 Dec 2004, 04:20
stolyar wrote:
Is [E!+Q!]! even?

(1) E=Q
(2) Q=2


[E!+Q!]! = (positive integer + positive integer)! = (integer >=2)! = always even
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 [#permalink] New post 16 Mar 2005, 21:01
Yeah... I see.
I have f--cked up myself. Sorry about it.

I am growing old... gettin' stupid
remember me in my golden days?
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Re: more fac [#permalink] New post 16 Mar 2005, 22:58
ChallengeMaker wrote:
stolyar wrote:
Is [E!+Q!]! even?
(1) E=Q
(2) Q=2

[E!+Q!]! = (positive integer + positive integer)! = (integer >=2)! = always even


yes. i was wondering about the question...............
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 [#permalink] New post 17 Mar 2005, 03:52
my analysis is

from statement 1


(1) E=Q then 0! + 0! = 2 Or 1! + 1! = 2 and so on so A is good


(2) Q=2 here we do not know what E is so if E = 1 0r 2 then u get 1+ 2 thats odd if E = 2 thne its even so statement 2 not suff

my answers to this is A
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 [#permalink] New post 17 Mar 2005, 08:41
Note the factorial sign outside of the bracket rxs.
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Re: more fac [#permalink] New post 17 Mar 2005, 17:31
MA wrote:
ChallengeMaker wrote:
stolyar wrote:
Is [E!+Q!]! even?
(1) E=Q
(2) Q=2

[E!+Q!]! = (positive integer + positive integer)! = (integer >=2)! = always even


yes. i was wondering about the question...............


Guys, the question doesnot need any more information to answer the question. try without statement i and statement ii.
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 [#permalink] New post 17 Mar 2005, 18:13
stolyar wrote:
Yeah... I see.
I have f--cked up myself. Sorry about it.

I am growing old... gettin' stupid
remember me in my golden days?


That is a great question stolyar! And great discussions and explantions. You should drop by more often!
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Answer [#permalink] New post 18 Mar 2005, 02:54
I will go with D.
At first i missed the outside factorial sign the moment i saw i knew all number greater than 2 will give even number. So even the factorials are O or 1 you have 2.
:)
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Re: more fac [#permalink] New post 18 Mar 2005, 05:04
stolyar wrote:
Is [E!+Q!]! even?

(1) E=Q
(2) Q=2


C.

unless Q & E are shown to be integers, they could otherwise be irrational.
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 [#permalink] New post 18 Mar 2005, 13:41
unless i am crazy you couldn't have factorials without using integers right?

Based on that assumption i throw in for D too.
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 [#permalink] New post 20 Mar 2005, 00:09
nice problem, D as well
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 [#permalink] New post 20 Mar 2005, 13:19
It should be (d) as any factorial other than 0 and 1 is even.
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 [#permalink] New post 20 Mar 2005, 14:47
i did not notice the outer most factorial so it should be D
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 [#permalink] New post 20 Mar 2005, 15:30
The outermost factorial makes all the difference.
  [#permalink] 20 Mar 2005, 15:30
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